This set of Theory of Machines Multiple Choice Questions & Answers (MCQs) focuses on “Worm Gears”.

1. When large gear reductions are needed _________ gears are used.

a) helical

b) spur

c) worm

d) bevel

View Answer

Explanation: Worm gears are used where large speed reductions are needed. The horizontal portion of the gear is called as a worm and the assembly of it with the gear is known as a worm gear. Worm can easily turn a gear but a gear cannot turn a worm because of the shallow angle on the worm.

2. The driven gear in the worm gear is a helical gear. True or false?

a) True

b) False

View Answer

Explanation: The helical gear is driven in the worm gear and the driving element is the screw or the worm. Worm gears are used in transmission of power between two non-parallel and non-intersecting shafts.

3. Which is of these is an advantage of worm gear?

a) It is expensive

b) Has high power losses and low transmission efficiency

c) Produce a lot of heat

d) Used for reducing speed and increasing torque

View Answer

Explanation: Reduction of speed and increasing the torque is an advantage of worm gear. The rest are the disadvantages of the worm gear. Worm gears are used in gate control mechanisms, hoisting machines, automobile steering mechanisms, lifts, conveyors and presses.

4. The distance between corresponding points on adjacent teeth measured along the direction of the axis is called ____________

a) joint line

b) normal link

c) axial pitch

d) lead

View Answer

Explanation: Axial pitch is the distance between corresponding points on adjacent teeth measured along the direction of the axis. The axial pitch of the worm gear is the same thing as the circular pitch of the helical gear.

5. The distance by which a helix advances along the axis of the gear for one turn around is called _____________

a) joint line

b) normal link

c) axial pitch

d) lead

View Answer

Explanation: Lead is the distance by which a helix advances along the axis of the gear for one turn around. The axial pitch is equal to the lead in a single helix and the axial pitch is one half of the lead in a double helix and so on.

6. The angle at which the teeth are inclined to the normal of the axis of rotation is called _______________

a) pitch angle

b) lead angle

c) normal angle

d) joint angle

View Answer

Explanation: Lead angle is the angle at which the teeth are inclined to the normal of the axis of rotation. Lead angle of the worm gear is same as the helix angle of the helical gear. Thus, Ψ2 = λ1.

7. What is the velocity ratio of worm gears?

a) (lπ)/d_{2}

b) (πd_{2})/l

c) l/(πd_{2})

d) d_{2}/(lπ)

View Answer

Explanation: Velocity ratio in worm gears is given as the ratio of the angle turned by the gear to the angle turned by the worm.

Thus, velocity ratio = (2l/d

_{2})/(2π) = l/(πd

_{2}).

8. What is the centre distance for the worm gear?

a) (m_{n}/2)(T_{1} cotλ_{1} – T_{2})

b) (m_{n}/2)(T_{2} cotλ_{1} + T_{1})

c) (m_{n}/2)(T_{2} cotλ_{1} – T_{1})

d) (m_{n}/2)(T_{1} cotλ_{1} + T_{2})

View Answer

Explanation: C = (m

_{n}/2)(T

_{1}cotλ

_{1}+ T

_{2})

This equation can be derived by using the formula for centre distance of a helical gear which is given as C = (m

_{n}/2) ((T

_{1}/cos Ψ

_{1}) + (T

_{2}/cos Ψ

_{2})

As, Ψ

_{2}= λ

_{1}, Ψ

_{1}= 90° – λ

_{1}.

9. What is the formula to calculate maximum efficiency of a worm gear?

a) (1+sinø)/(1-sinø)

b) (1-sinø)/(1+sinø)

c) (tan(λ_{1}-ø))/tan λ_{1}

d) (tan(λ_{1}+ø))/tan λ_{1}

View Answer

Explanation: The maximum efficiency of the worm gear is given to be (1-sinø)/(1+sinø), whereas the efficiency of the worm gear is given as (tan(λ

_{1}-ø))/tan λ

_{1}

10. Find the helix angle of the worm if the pitch of the worm gear is 12 mm and the pitch diameter is 50 mm.

a) 8.687°

b) 11.231°

c) 9.212°

d) 10.319°

View Answer

Explanation: tan λ

_{1}= Lead / Pitch circumference = 2p/πd1 = 24/50π = 0.1528

λ

_{1}= 8.687°.

11. Find the speed of the gear if the worm is a three start worm rotating at 500 rpm. The gear has 20 teeth.

a) 125 rpm

b) 100 rpm

c) 75 rpm

d) 50 rpm

View Answer

Explanation: N

_{1}/N

_{2}= T

_{2}/T

_{1}

500/N

_{2}= 20/3

N

_{2}= 75 rpm

Thus, the gear rotates at a speed of 75 rpm.

12. For a two start worm gear having a pitch of 20 mm and a lead angle 12°, find the centre distance if the larger gear has 25 teeth.

a) 148.22 mm

b) 124.93 mm

c) 121.19 mm

d) 109.53 mm

View Answer

Explanation: C = (m

_{n}/2)(T

_{1}cotλ

_{1}+ T

_{2})

Therefore, C = (p

_{n}/2π)(T

_{1}cotλ

_{1}+ T

_{2}) = 109.53 mm.

13. Calculate the lead angle of the worm gear for maximum efficiency if θ = 90° and the coefficient of friction is 0.05.

a) 48.21°

b) 42.23°

c) 43.57°

d) 46.43°

View Answer

Explanation: µ = 0.05°; ø = tan

^{-1}(0.05) = 2.862°; θ = 90°

For maximum efficiency, Ψ

_{1}= (θ+ ø)/2 = 92.862/2 = 46.43°

Ψ

_{1}= 90° – λ

_{1}= 46.43°

λ

_{1}= 90° – 46.43° = 43.57°.

14. Find the maximum efficiency if the lead angle is given to be 10° and the coefficient of friction is 0.07.

a) 79.82%

b) 72.23%

c) 76.29%

d) 70.72%

View Answer

Explanation: λ

_{1}= 10°, ø = tan

^{-1}(0.07) = 4°

Efficiency = tan(λ

_{1})/ tan(λ

_{1}+ ø) = 0.7072 = 70.72%.

15. Calculate the maximum efficiency of the worm gears which have a friction angle of 0.06.

a) 88.71%

b) 83.23%

c) 89.91%

d) 86.49%

View Answer

Explanation: ø = tan

^{-1}(0.06) = 3.43°

Maximum efficiency = (1-sin ø)/(1+sin ø) = 0.8871 = 88.71%.

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